Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>A sector of a circle of radius 1 with angle α is bent to form a cone, where \( r = 1 - \dfrac{\alpha}{2\pi} \). The value of \( r \) for which volume is maximum (when α is variable):</p>
<p>\( \dfrac{\sqrt{2}}{3} \)</p>
<p>\( \dfrac{2}{\sqrt{3}} \)</p>
<p>\( \sqrt{\dfrac{2}{3}} \)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: The arc length of the sector (α·1 = α) becomes the circumference of the cone's base (2πr), establishing the constraint 2πr = α. The volume V = (1/3)πr²h must be maximized using the given relationship r = 1 - α/(2π) to express V as a function of a single variable.
<p><strong>Step 1:</strong> When the sector is bent into a cone, the arc length becomes the base circumference:</p><p>Arc length = α · 1 = α</p><p>Base circumference = 2πr</p><p>Therefore: α = 2πr (which is consistent with the given r = 1 - α/(2π))</p><p><strong>Step 2:</strong> The slant height of the cone is 1. Using Pythagorean theorem:</p><p>h² + r² = 1²</p><p>h = √(1 - r²)</p><p><strong>Step 3:</strong> Volume of cone:</p><p>V = (1/3)πr²h = (1/3)πr²√(1 - r²)</p><p><strong>Step 4:</strong> To maximize, take dV/dr:</p><p>dV/dr = (1/3)π[2r√(1 - r²) + r² · (-r/√(1 - r²))]</p><p>= (1/3)π[2r√(1 - r²) - r³/√(1 - r²)]</p><p>= (1/3)π · r/√(1 - r²) · [2(1 - r²) - r²]</p><p>= (1/3)π · r/√(1 - r²) · (2 - 3r²)</p><p><strong>Step 5:</strong> Setting dV/dr = 0 (for r > 0):</p><p>2 - 3r² = 0</p><p>r² = 2/3</p><p>r = √(2/3) = √6/3</p><p><strong>Step 6:</strong> Verify this is a maximum by checking d²V/dr² < 0 or noting that V = 0 at r = 0 and r = 1, so the critical point is a maximum.</p><p>∴ Answer: C (r = √6/3 or equivalently √(2/3))</p>
Correct Answer: C